# If the square root of two is irrational, why can it be created by dividing two numbers?

I'm doing pre-calculus and got a bit caught on this question… I looked online and it said that a number is rational if it can be the quotient of two integers. So I did this: I know that it is irrational because if this was actually rational, mathematicians would have figured out. So my question is, why is the square root of two irrational?

• You've found an approximation that is accurate to all the digits displayed. However, your quantity terminates at the ...237, while the actual sqrt(2) goes on forever. – QuantumFool Oct 25 '16 at 6:45
• You only have an approximation of $\sqrt{2}$. $\sqrt{2}$ is a real number, hence has an infinite decimal expansion, which is non-repeating non-terminating by the fact that it is irrational. As for the reason why it is irrational, I want you to look up a famous proof by contradiction by Euclid (somebody should post it as an answer below in future). – астон вілла олоф мэллбэрг Oct 25 '16 at 6:46
• @астонвіллаолофмэллбэрг does that mean all irrational numbers are infinite? – Jake Wickham Oct 25 '16 at 7:49
• @JakeWickham It is good that you have asked this question. The answer is : there is a difference between having an infinite decimal expansion and being infinite. The reason is very simple : The sum of infinitely many numbers can possibly be finite, and that is what is happening. $1.414213 = 1 + \frac{4}{10} + \frac{1}{100} + \frac{4}{1000} + \ldots$. You are adding infinitely many numbers, but the sum remains finite. To be infinite is , in some sense, to be larger in absolute value than any natural number, you could say. – астон вілла олоф мэллбэрг Oct 25 '16 at 7:54
• Don't take calculator output literally. Calculator outputs are approximate; your calculator might give a terminating decimal result for $\sqrt{2}$, but that's not the actual square root. – user2357112 supports Monica Oct 25 '16 at 19:30

Squaring the number you posted:

$$1.41421356237 \cdot 1.41421356237 = 1.9999999999912458800169$$

so the result is not $2$, so the number is not $\sqrt{2}$ (but rather a rational approximation thereof).

• This calculation, when worked out, might be longer by some measures than the proof that x^2=2 has no rational solutions (plus application of that to the particular x). – zyx Oct 25 '16 at 18:16
• @zyx Realistically, you can just punch it into a calculator with more digits. – user253751 Oct 26 '16 at 4:57
• @zyx: (and @immibis) Since the product of the two right-most digits is $(7)(7)=49$ (oops, don't look here), it follows that the right-most digit of the resulting base-10 numeral is $9,$ which means the result can't be $2$ (even when written in its alternate form as $1.999\ldots).$ – Dave L. Renfro Oct 26 '16 at 14:23

This is the usual proof by contradiction that $\sqrt2$ is irrational (often misattributed to Euclid):

Suppose $\sqrt2$ is rational.
Then $\sqrt2=\dfrac mn$ where $m,n\in\Bbb N$ and $\gcd(m,n)=1$.
Squaring, $2=m^2/n^2\implies2n^2=m^2\implies m=2k$ for some $k\in\Bbb N$ ($m$ is even).
Then $2n^2=4k^2\implies n^2=2k^2$, so $n$ must be even.
But now we see that 2 divides both $m$ and $n$, which contradicts the initial assumption that $\gcd(m,n)=1$.
Hence $\sqrt2$ is irrational.

Your attempt at showing the rationality of $\sqrt2$ fails because we can take your fraction through this argument and arrive at absurdities (e.g. $2n^2=m^2$, which is false by direct calculation).

However, we can get arbitrarily close to $\sqrt2$ with rational numbers, and good approximations have been known for centuries, such as $99/70$ and $577/408$. These stem from the continued fraction expansion of $\sqrt2$ as $[1;2,2,2,2,\dots]$ and are better than your decimal digit-based expansions in the sense that they have the smallest denominator for a given accuracy.

• Note that gcd(m, n) = 1 is necessary for this to work. If we had gcd(m, n) = a for some a > 1, then we could simply reduce the fraction to its simplest form by dividing top and bottom by a and then proceed as before. – paolo Oct 25 '16 at 12:11
• Book X of Euclid's Elements has only 115 propositions. And while this may be a reasonable translation of Euclid's argument (wherever it is found) into modern number theory, note that it is far different from the actual argument that Euclid provided, which was geometric in nature. – Paul Sinclair Oct 25 '16 at 16:23
• I don't quite see how this proves it. Surely, n^2 may be any real number, not just an integer, so n does not need to be even. – DeadMG Oct 25 '16 at 23:13
• It's explicitly stated that $m,n\in \mathbb{N}$. $n$ and therefore $n^2$ by definition cannot be just "any real number", but in fact are positive integer and positive integer square, respectively. – Nij Oct 26 '16 at 4:41
• This doesn't answer the question. The question wasn't "find a proof that $\sqrt2$ is irrational". It was "what exactly is wrong with my proof that $\sqrt2$ is rational". – msh210 Oct 26 '16 at 21:27

The square root of two cannot exactly be written out on a computer screen in decimal notation – the digits extend forever. You're looking at an approximation accurate to a lot of digits. The approximation itself, however, is rational, as you have demonstrated. If you expanded it out to more digits, however, you'd realize that your quantity has zeros extending out to infinity, while $\sqrt{2}$ has more digits. Therefore your rational number does not contradict the proof given by the mathematical community that $\sqrt{2}$ is irrational.

Note: just because a number is irrational doesn't mean it can't be arbitrarily well approximated by rational numbers – every number, rational and irrational can, and that property is known as "the rational numbers are dense in the reals." For instance, for $\sqrt{2}$: \begin{align} \frac{1}{1}, \frac{14}{10}, \frac{141}{100}, \frac{1414}{1000}, \frac{14142}{10000}, ... \end{align} get arbitrarily close to $\sqrt{2}$, but none of them are equal to it.

Related topic: if you're interested in rational numbers which quickly converge to irrational numbers, check out continued fractions. They're really cool. Interesting numbers like $\sqrt{2}$ often have continued fraction expansions with closed-form coefficients – $\sqrt{2}$ can be represented by a continued fraction with first coefficient 1 and all following coefficients 2. You can do all sorts of weird things with this, like show that the Golden Ratio (which has a continued fraction expansion of all 1's) is (one of) the irrational number(s) which takes the longest to converge to, or in more plain concepts, the rational numbers used to approximate it require a really large denominator to do a good job.

• Was going to write that until I read your answer.. Very good ! +1 :) – GameDeveloper Oct 27 '16 at 9:15
• You can't write 1/7 in finite decimal notation either, it periodically repeats itself. – Erbureth says Reinstate Monica Oct 30 '16 at 20:44
• @Erbureth Sure. Just because a number cannot be written in a finite number of digits base-10 doesn't make it irrational. However, the converse is false: a number that can be written in a finite number of decimal digits is not irrational. The OP thought he had written $\sqrt{2}$ down in finitely many digits and therefore it was rational; I was explaining how that's only an approximation and therefore the irrationality of $\sqrt{2}$ remains intact. This isn't a proof that $\sqrt{2}$ is irrational, only that the OP's argument for why it should be rational is false. – QuantumFool Oct 30 '16 at 21:09
• Right, sorry, I misunderstood your answer. – Erbureth says Reinstate Monica Oct 30 '16 at 22:24

"Does that mean all irrational numbers are infinite?"

The irrational numbers are exactly those numbers that require infinitely many digits right of the radix point, no matter which base you are using. (Thanks Beanluc: Normally we say "decimal point" but that would obviously wrong in bases other than base 10, so the generic term is "radix point". )

A number like 1/3 needs infinitely many digits in base 2, 10, 16 but not in base 3, 6, 9, 12. $2^{1/2}$ needs infinitely many digits in every base.

There's nothing so mysterious: $$\left(\frac{141421356237}{10^{11}}\right)^{\!2}= \frac{19999999999912458800169}{10^{22}}\ne 2$$ Thus $\dfrac{141421356237}{10^{11}}$ is not $\sqrt{2}$.
Without computations: $141421356237$ is odd, so its square is odd too. Hence the fraction cannot simplify to an even integer.